NCERT Solutions for Class 10 Maths Unit 8

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NCERT Solutions for Class 10 Maths Unit 8

Introduction to Trigonometry Class 10

Unit 8 Introduction to Trigonometry Exercise 8.1, 8.2, 8.3, 8.4 Solutions

Exercise 8.1 : Solutions of Questions on Page Number : 181 Q1 :

In ABC right angled at B, AB = 24 cm, BC = 7 m. Determine (i) sin A, cos A (ii) sin C, cos C

Answer : Applying Pythagoras theorem for ABC, we obtain AC2 = AB2 + BC2 = (24 cm)2 + (7 cm)2 = (576 + 49) cm2 = 625 cm2

AC =

cm = 25 cm

(i) sin A =

cos A = (ii)





sin C =

cos C =

Q2 : In the given figure find tan P - cot R

Answer : Applying Pythagoras theorem for PQR, we obtain PR2 = PQ2 + QR2 (13 cm)2 = (12 cm)2 + QR2 169 cm2 = 144 cm2 + QR2 25 cm2 = QR2 QR = 5 cm





tan P - cot R =

Q3 :

If sin A = , calculate cos A and tan A. Answer : Let ABC be a right-angled triangle, right-angled at point B.

Given that,





Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain AC2 = AB2 + BC2 (4k)2 = AB2 + (3k)2 16k 2 - 9k 2 = AB2 7k 2 = AB2 AB =

Q4 : Given 15 cot A = 8. Find sin A and sec A Answer : Consider a right-angled triangle, right-angled at B.

It is given that,





cot A = Let AB be 8k.Therefore, BC will be 15k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain AC2 = AB2 + BC2 = (8k)2 + (15k)2 = 64k2 + 225k2 = 289k2 AC = 17k

Q5 : Given sec = , calculate all other trigonometric ratios. Answer : Consider a right-angle triangle ABC, right-angled at point B.





If AC is 13k, AB will be 12k, where k is a positive integer. Applying Pythagoras theorem in ABC, we obtain (AC)2 = (AB)2 + (BC)2 (13k)2 = (12k)2 + (BC)2 169k2 = 144k2 + BC2 25k2 = BC2 BC = 5k

Q6 :

If A and B are acute angles such that cos A = cos B, then show that A = B.

Answer :

Let us consider a triangle ABC in which CD AB.

It is given that cos A = cos B





... (1)

We have to prove A = B. To prove this, let us extend AC to P such that BC = CP.

From equation (1), we obtain

By using the converse of B.P.T, CD||BP

ACD = CPB (Corresponding angles) ... (3) And, BCD = CBP (Alternate interior angles) ... (4)

By construction, we have BC = CP.

obtainCBP = CPB (Angle opposite to equal sides of a triangle) ... (5) From equations (3), (4), and (5), we ACD = BCD ... (6)

ACD = BCD [Using equation (6)]

CDA = CDB [Both 90?]

Therefore, the remaining angles should be equal.

CAD = CBD A = B

Alternatively,

Let us consider a triangle ABC in which CD AB.





It is given that, cos A = cos B

Let

AD = k BD ... (1)

And, AC = k BC ... (2)

Using Pythagoras theorem for triangles CAD and CBD, we obtain

CD2 = AC2 - AD2 ... (3)

And, CD2 = BC2 - BD2 ... (4)

From equations (3) and (4), we obtain

AC2 - AD2 = BC2

(k BC)2 - (k BD)2 = BC2 - BD2 k2 (BC2 - BD2) = BC2 - BD2 k2 = 1 k=1

- BD2

A = B(Angles opposite to equal sides of a triangle)

Q7 : If cot = , evaluate

(i)

(ii) cot2



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